Question

How do you find the discontinuities for `y=(4x)/(x^3-9x)`?

Answer 1

Factor the denominator to get:

`y = (4x)/(x(x-3)(x+3)) = 4/((x-3)(x+3))`

with exclusion `x != 0`

This has a removable discontinuity at `x=0` and simple poles at `x=+-3`

Explanation:

`y = f(x) = (4x)/(x(x-3)(x+3)) = 4/((x-3)(x+3))*x/x`

`= 4/((x-3)(x+3))` with exclusion `x != 0`

`f(0)` is undefined since `0/0` is undefined,

So `f(x)` has a discontinuity at `x=0`.

This is a removable discontinuity as we can simply define:

`g(0) = lim_(x->0) f(x) = -4/9`

`g(x) = f(x)` when `x != 0`

`f(-3)` and `f(3)` are simple pole discontinuities, where the denominator is zero and the numerator is non-zero.

graph{4x/(x(x-3)(x+3)) [-10, 10, -5, 5]}